Hello,
I am trying to upload data I have on a Google Spreadsheet within R to
perform some analysis. I regularly update such data and need to perform
data analysis in the quickiest possible way - i.e. without need to publish
the data, so I was wondering how to make work this piece of code (source
h
We have been able to reproduce the reported issue on another Linux
system: Fedora 19, and the solution was quite simple:
The deSolve package must always to be loaded *before* loading the shared
library of the compiled model.
Thomas
__
R-help@r-project
On Thu, 7 Nov 2013, sv.j...@yahoo.ca wrote:
Hello,
I have count data for 4 groups, 2 of which have a large number of zeroes
and are overdispersed, and the other 2 underdispersed with no zeroes.
Are you sure that it's really underdispersion in addition to the lack of
zeros? It could also be
Hello,
I am trying to plot ADMIXTURE output in R. My output is two-dimensional
(two columns, 190 rows). These rows represent individuals, and groups of
them (e.g., 1-12) represent populations. What I'm hoping for is a plot with
eleven different colours, basically like a STRUCTURE plot.
#This code
Hello,
I have count data for 4 groups, 2 of which have a
large number of zeroes and are overdispersed, and the other 2
underdispersed with no zeroes. I have two questions about model fitting,
which I am quite new to, and have been using mostly the pscl package.
1 - How do I deal with underdisp
Hi
Take a look at current.transform() (the grid.locator() function shows an
example use)
Paul
On 11/06/13 13:35, James Price wrote:
I'm trying to do some post-plot manipulation of some lattice
graphics, in which I need to get the absolute viewport locations on
the plotting device. So for exa
Hello,
I am fitting curves in drc to a Log-logistic 4-parameter model.
My experiment is looking at fall and spring applications of a herbicide on
wild oat at 5 different locations. A preliminary mixed model ANOVA showed
that timing was significant. I am running a dose response curve for each
app
If you already have the order stored in a list or so:
For example:
dat1 <- as.data.frame(mat,stringsAsFactors=FALSE)
lst1 <- list(c("OF","ON"), c("US","UK", "WW","BO","BR","CA"),
c("P2","P3","P1"),c("S2","S1","S3"))
dat1[] <- lapply(seq_along(lst1),function(i)
factor(dat1[,i],levels=lst1[[i]]))
It appears to be legitimate to include multi-level categorical and continuous
variables in defining the model for earth (e.g. y ~ cat + cont1 + cont2) but is
it also then possible use categoricals in the predict method using the earth
result? I tried but it returns an error which is not very inf
Hi,
Not sure whether this helps:
dat1 <- as.data.frame(mat,stringsAsFactors=FALSE)
dat1$c4 <- factor(dat1$c4,levels=c("OF","ON"))
dat1$c1 <- factor(dat1$c1,levels=c("US","UK","WW","BO","BR","CA"))
dat1$c2 <- factor(dat1$c2, levels=c("P2","P3","P1"))
dat1$c3 <- factor(dat1$c3, levels=c("S2","S1"
You might need to implement it as a nested pair of for loops using rbind.
In essence iterate over the rows in df1 and each time find the matching
row in df2. If none is found then add the df1 row by itself to the
result. If one is then remove it from df2 and rbind both of them. Once
done just me
On Nov 7, 2013, at 12:55 PM, Barrett Gady wrote:
> does anyone know how to resolve Error: .onLoad failed in loadNamespace() for
> ‘xlsxjars’
>
> I can’t seem to get this package to load on my windows 7 install of R3.02
It appears that you did not install all the depndencies for package xls
On 13-11-07 6:39 PM, Filippo wrote:
Hi,
I'm having strange differences between the R function prod ad the F90
function product.
Processing the same vector C (see attachment). I get 2 different results:
prod(C) = 1.069678e-307
testProduct(C) = 0
where testProd is the following wrapping function:
Fortune !
Bert
Sent from my iPhone -- please excuse typos.
> On Nov 7, 2013, at 3:48 PM, Rolf Turner wrote:
>
>> On 11/08/13 12:39, Filippo wrote:
>> Hi,
>> I'm having strange differences between the R function prod ad the F90
>> function product.
>> Processing the same vector C (see attachme
Dear all,
I would like to merge two dataframes using two conditions. For example, if
I have the dataframes df1 and df2:
> (df1 <- data.frame(var1=c("a","b","d","e","g"), var2=c(25,14,53,26,84),
infodf1=c(1,1,1,1,1)))
var1 var2 infodf1
1a 25 1
2b 14 1
3d 53 1
On 11/08/13 12:39, Filippo wrote:
Hi,
I'm having strange differences between the R function prod ad the F90
function product.
Processing the same vector C (see attachment). I get 2 different results:
prod(C) = 1.069678e-307
testProduct(C) = 0
If you are worried about the difference between
Hi,
I'm having strange differences between the R function prod ad the F90
function product.
Processing the same vector C (see attachment). I get 2 different results:
prod(C) = 1.069678e-307
testProduct(C) = 0
where testProd is the following wrapping function:
testProd <- function(x) {
retu
Martin Turcotte gmail.com> writes:
> Dear list,
> I am trying to use MuMIn to compare all possible mixed models
> using the dredge function on binomial data but I
> am getting an error message that I cannot decode. This error
> only occurs when I use glmer. When I use an lmer
> analysis on a d
does anyone know how to resolve Error: .onLoad failed in loadNamespace() for
âxlsxjarsâ
I canât seem to get this package to load on my windows 7 install of R3.02
Sent from Windows Mail
[[alternative HTML version deleted]]
__
R-help
Yep, that solves the problem and is much cleaner than what I was trying to do.
KWS
On Nov 7, 2013, at 10:01 AM, arun wrote:
> Hi,
>
> May be this is what you wanted:
> mat1 <- matrix(1:4,2,2)
> results.norm <- mat1/rowSums(mat1)
>
> A.K.
>
>
>
>
> On Thursday, November 7, 2013 12:27 PM, K
Hi,
You can check the dataset for any patterns that are unique for the datetime
values.
For example:
vec1 <- c(as.character(Sys.time()), format(Sys.time(), "%a %b %d %X %Y
%Z"),4324343,format(Sys.time(),"%m-%d-%Y %H:%M:%S"),"aZZCRDDS")
vec1[grepl(":",vec1)]
#[1] "2013-11-07 15:04:27"
If you use GCV smoothness selection then, in the Gaussian case, the key
assumptions are constant variance and independence. As with linear
modelling, the normality assumption only comes in when you want to find
confidence intervals or p-values. (The GM Thm does not require normality
btw. but I do
In the absence of any description of the computers themselves, it's hard to
say, but there's a good chance that there is some object in your working
environment that's being accessed by your function -- and that object exists
on only one of the machines.
I would suggest you use debug() or browser
So is there a way to make the subsetting behave as expected?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Doran, Harold
Sent: 07 November 2013 16:52
To: 'Eckstädt, Elisabeth'; r-help@R-project.org
Subject: Re: [R] strange behav
Dear list,
I am trying to use MuMIn to compare all possible mixed models using the dredge
function on binomial data but I am getting an error message that I cannot
decode. This error only occurs when I use glmer. When I use an lmer analysis on
a different response variable every works great.
... and actually, since u can be assumed to be of the form shown,
v <-do.call(expand.grid, split(rep(u,len),rep(u,e=len)))
should do.
-- Bert
On Thu, Nov 7, 2013 at 10:06 AM, Bert Gunter wrote:
> Well, you can create the expand.grid data frame programmatically via:
>
> u <- 1:3
> len <- lengt
Hélène,
thanks for spotting this! This is a bug in "AER". I had just tested the
new diagnostics for regressions with 1 endogenous variable and hence
never noticed the problem. But if there are > 1 endogenous variables, the
df used in ivreg() (and hence the associated p-values) are too large.
Well, you can create the expand.grid data frame programmatically via:
u <- 1:3
len <- length(u)
v <-do.call(expand.grid, split(rep(u,len),rep(seq_len(len),e=len)))
And then you can use unique.array to get the unique rows after the sort:
unique(t(apply(v,1,sort)))
However, I agree with your sent
2013/11/6 Michael Friendly :
> On 11/4/2013 10:45 AM, Sergio Fonda wrote:
>>
>> Hi,
>> I'm not able to get information about the following question:
>>
>> is the variables standardization a default option in manova() (stats
>> package)?
>> Or if you want to compare variables with different units or
Ken Knoblauch inserm.fr> writes:
>
> Roland Deutsch tuwien.ac.at> writes:
> > in my research I frequently work with binomial
> response models, which
> > are of course part of the generalized linear
> models. While I do use
> > common link functions such as the logit, probit
> and cloglog,
On 07-Nov-2013 13:38:29 Konstantin Tretiakov wrote:
> Hello!
>
> I need to obtain all possible combinations with replacement when
> order is not important.
> E.g. I have a population x{1,2,3}.
> So I can get (choose(3+3-1,3)=) 10 combinations from this population
> with 'size=3'.
> How can I get a
Yes, what I would do is write a function around the all.equal() comparisons on
the data to see which values are "nearly" identical to the ones desired I the
subset. If TRUE, I would retain those values and discard the others.
-Original Message-
From: Simon Hayward [mailto:simon.hayw...@
Yes, but notice that
man2[3,] == .3
[1] FALSE
This is because of issues of machine precision when dealing with floating
points, not a problem in R. Comparisons for nearly equivalent numbers are done
using all.equal() as shown below.
> all.equal(man2[3,], .3)
[1] TRUE
-Original Message
Is this what you want?
f <- function (x, m) combn(x + m - 1, m) - seq_len(m) + 1
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Konstantin Tretiakov
> Sent: Thurs
Hello
I am running exactly the same code on two different computers. In PC1 of the
the code is working fine, whereas in PC2 I get this error
Error in HP.LUT[, 1] : incorrect number of dimensions
HP.LUT is an object within a function computed as HP.LUT <- which(HP ==1,
arr.in=TRUE
Roland Deutsch tuwien.ac.at> writes:
> in my research I frequently work with binomial
response models, which
> are of course part of the generalized linear
models. While I do use
> common link functions such as the logit, probit
and cloglog, I often
> have the need of invoking the lesser-kno
sorry i missed that. install.packages('foreign') solved the problem,
thank you!! :)
On Thu, Nov 7, 2013 at 10:41 AM, Prof Brian Ripley wrote:
> See the posting guide: your version of 'foreign' is not current:
> http://cran.r-project.org/web/packages/foreign/index.html .
>
> Please update yo
See the posting guide: your version of 'foreign' is not current:
http://cran.r-project.org/web/packages/foreign/index.html .
Please update your packages and try again. (This looks very like a bug
in recently contributed code that has already been fixed.)
On 07/11/2013 13:40, Anthony Damico w
R FAQ 7.31.
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
On Thu, Nov 7, 2013 at 8:36 AM, Eckstädt, Elisabeth
wrote:
> Hello everyone,
> I am experiencing a unfathomable benaviour of "subset" on a data.frame. This
> is a minimal reproducable e
Hi ,
I would like to check if a column contains datetime values. (column may
contain any datetime format or just numbers or string). Request your help
--
View this message in context:
http://r.789695.n4.nabble.com/Checking-datetime-value-tp4679963.html
Sent from the R help mailing list archiv
Hello everyone,
I am experiencing a unfathomable benaviour of "subset" on a data.frame. This is
a minimal reproducable example. The data.frame cosists only of one column,
which contains 10 ascending values (from 0.1 to 1). Subsetting for 0.1 is
working (gives me one row), subsetting for 0.3 give
Hello!
I need to obtain all possible combinations with replacement when order is
not important.
E.g. I have a population x{1,2,3}.
So I can get (choose(3+3-1,3)=) 10 combinations from this population with
'size=3'.
How can I get a list of all that combinations?
I have tried 'expand.grid()' and ma
Hello,
I'm new to R and I'm currently learning to use package AER, which is
extremely comprehensive and useful. I have one question related to the
diagnostics after ivreg: if I understood well, the Sargan test provided
states that the statistic should follow a Chi squared of degrees of freedom
equa
Dear R Help,
in my research I frequently work with binomial response models, which
are of course part of the generalized linear models. While I do use
common link functions such as the logit, probit and cloglog, I often
have the need of invoking the lesser-known Complementary Log link
(Walter
First of all sorry for my bad english but its not my native language.
I am working on a paper on Portfolio Optimization with Markowitz and Lower
Partial Moments.
I want to compare the returns of the minimum variance portfolios from booth
methods. First of all i have an in-sample multivariate tim
Use KernSmooth (one of the recommended packages that are included in R
distribution). E.g.,
> library(KernSmooth)
KernSmooth 2.23 loaded
Copyright M. P. Wand 1997-2009
> x <- seq(0, 1, length=201)
> y <- 4 * cos(2*pi*x) + rnorm(x)
> f <- locpoly(x, y, degree=0, kernel="epan", bandwidth=.1)
> plo
a_lampei uni-tuebingen.de> writes:
>
> Dear all,
> I have a problem with interactions in lmer. I have 2 factors (garden and
> gebiet) which interact, plus one other variable (home),
> dataframe arr. When
> I put:
> /
> lmer (biomass ~ home + garden:gebiet + ( 1|Block), data = arr)/
>
> it wri
this file
http://www.electionstudies.org/studypages/data/anes_mergedfile_1992to1997/anes_mergedfile_1992to1997_dta.zip
can be downloaded after free registration on this page
http://electionstudies.org/studypages/download/registration_form.php
imports properly in windows R x64 3.0.1 but ca
Here is a start. I was wondering how long it would take to at least
substitute 800 different patterns into 4M vectors. Here is my test.
It took longer (99 sec) to create the test data than to do the
substitutes (52 secs). Now some variations on this can provide the
other information that you are
You already asked this on StackOverflow.
The answer remains the same, pretty much what David W. wrote: this is
not a question about fitting lines to data. You need to step back and think
about what message you want to deliver to those who will view your graph,
and what the meaning of your data
Dear all,
I have a problem with interactions in lmer. I have 2 factors (garden and
gebiet) which interact, plus one other variable (home), dataframe arr. When
I put:
/
lmer (biomass ~ home + garden:gebiet + ( 1|Block), data = arr)/
it writes:
/Error in lme4::lFormula(formula = biomass ~ home + ga
Dear Alfonso,
in a RE model you do not explicitly estimate every single individual effect,
but only the variance of the distribution they have been "drawn from". Hence
the only (pointwise) residual you can estimate ex-post is the composite one:
i.e., the sum.
Best,
Giovanni
Giovanni Millo, P
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